Can a charged capacitor mathematically be represented as uncharged capacitor connected in series with battery of opposing emf ?
Please explain with an example
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$\begingroup$ It can be done to find the heat generated but not to find final charges $\endgroup$– user1825567Commented Apr 21, 2016 at 7:58
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$\begingroup$ I think the argument should be "Can a charged capacitor mathematically be represented as a battery of opposing emf ?" $\endgroup$– jimCommented Apr 21, 2016 at 10:55
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1 Answer
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The two networks will be equivalent if the e.m.f in the second network is equal to $Q/C$ in the first. Then you will get two networks with same initial voltage across their terminals and same impedance too (since $z_{voltage source}=0$), so they are equivalent.
As you are asking for an example, consider the simple R-C series circuit, first with a charged capacitor and no generator, then with an uncharged capacitor and an e.m.f.
Applying KVL at an instant $t$ until which a charge $q$ has passed through the circuit, in the first one we get:$$\dfrac{Q-q}{C}=Ri$$
which is a differential equation in $q$.
In the second:$$E=Ri+\dfrac{q}{C}$$but $E=\dfrac{Q}{C}$, then:$$\dfrac{Q-q}{C}=Ri$$ so it is the same differential equation, and will give the same equation for in both circuits.
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$\begingroup$ sir,so can this method be used to find charge flown in the circuit?i thought that the charge flown can be added to the initial charge on capacitors to find final charge of the system and hence easily find the heat generated.can it be done sir? $\endgroup$ Commented Apr 21, 2016 at 12:03
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$\begingroup$ @AjaySabarish the charge flowing is the same in both circuits, so is the heat generated. $\endgroup$– TofiCommented Apr 21, 2016 at 12:09